Limits and Continuity Test 42

Result

Limits and Continuity Test 42

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Weekly Quiz Competition

Question 1
1 / -0

$$lim_{n\to \infty} \Sigma^n_{r=1} \dfrac{\pi}{n} sin(\dfrac{\pi r}{n})$$ is equal to

A
$$1$$

B
$$2$$

C
$$3$$

D
$$4$$

Solution
Question 2
1 / -0

Evaluate : $$\displaystyle\lim _{ x\rightarrow 0 }{ \left( \dfrac { { e }^{ x\ell n\left( { 3 }^{ x }-1 \right) }-\left( { 3 }^{ x }-1 \right) ^{ x }\sin { x } }{ { e }^{ x\ell nx } } \right) } $$ is equal to

A
$$\dfrac{1}{e}\ell n3$$

B
$$e\ \ell n\ 3$$

C
$$3$$

D
$$\dfrac{1}{3}$$

Solution
Question 3
1 / -0

$$\underset{x \rightarrow 0}{Lt} \dfrac{sin x - x + x^3 / 6}{x^5}$$ =

A
$$\dfrac{1}{120}$$

B
$$\dfrac{1}{110}$$

C
$$\dfrac{1}{100}$$

D
none of these

Solution
Question 4
1 / -0

$$\displaystyle \lim_{x\rightarrow0 }{\dfrac{(\cos\alpha)^{x}-(\sin\alpha)^{x}-\cos 2\alpha}{(x-4)}}, \alpha\in \left(0, \dfrac{\pi}{2}\right)$$ is equal to

A
$$\cos^{4}\alpha.\log(\cos\alpha)-\sin^{4}\alpha.\log(\sin\alpha)$$

B
$$\sin^{4}\alpha.\log(\cos\alpha)-\cos^{4}\alpha.\log(\sin\alpha)$$

C
$$\sin^{4}\alpha.\log(\cos\alpha)+\cos^{4}\alpha.\log(\sin\alpha)$$

D
$$None\ of\ the \ above$$

Solution
Question 5
1 / -0

evaluate$$ \underset { x\rightarrow 0 }{ lim } \frac { x-\int _{ 0 }^{ x }{ { cost }^{ 2 }dt } }{ { x }^{ 3 }-6x } $$

A
$$3$$

B
$$-1$$

C
$$0$$

D
$$1$$

Solution
Question 6
1 / -0

$$\underset { x\rightarrow 0 }{ Lt } \cfrac {tanx-x}{x^2tanx}$$ equals:

A
1

B
1/2

C
1/3

D
None of these

Solution
Question 7
1 / -0

$$\underset { x\rightarrow \pi/2 }{ lim } \left(\dfrac{cosec x-1}{cot^2x}\right)= $$

A
$$0$$

B
$$-\dfrac{1}{2}$$

C
$$\dfrac{1}{2}$$

D
$$1$$

Solution
Question 8
1 / -0

$$\displaystyle\lim_{x\rightarrow \infty}\left(\dfrac{x+1}{2x+1}\right)^{x^2}$$ equals?

A
$$0$$

B
e

C
$$1$$

D
$$\infty$$

Solution
Question 9
1 / -0

$$\underset{x \rightarrow \infty}{lim} \dfrac{2 \tan^{-1} x}{\pi}$$ equals $$e^L$$ then $$L$$ is equal to

A
$$\dfrac{2}{\pi}$$

B
$$-\dfrac{2}{\pi}$$

C
$$-\dfrac{\pi}{2}$$

D
$$1$$

Solution
Question 10
1 / -0

$$\displaystyle \lim _{ x\rightarrow x/2 } \dfrac { \left[ 1-\tan { \left( \dfrac { x }{ 2 } \right) } \right] \left[ 1-\sin { x } \right] }{ \left[ 1+\tan { \left( \dfrac { x }{ 2 } \right) } \right] \left[ \pi -2x \right] ^{ 3 } } $$ is

A
$$\dfrac {1}{8}$$

B
$$0$$

C
$$\dfrac {1}{32}$$

D
$$\infty$$

Solution